
%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[10pt]{article} 

\input{wang_preamble.tex}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{titling}
\setlength{\droptitle}{-2cm}   % This is your set screw

%%文档的题目、作者与日期
\author{王立庆（2019级数学与应用数学1班）}
\title{数量金融实验 - 第3章课堂练习}
%\date{\vspace{-3ex}}
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{2022 年 9 月 8 日}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %1
设 $\{M_n,n\ge 0\}$ 是一维随机游动。则 $\{M_n, n\in [0,10]\}$ 一共有多少不同的样本路径？
\begin{enumerate}
\item[A.]  256. 
\item[B.]  512. 
\item[C.]  1024. 
\item[D.]  2048. 
\end{enumerate}

{\color{red}
解答：C. 由一维随机游动的定义，$M_0=0$. 每一步可以向左或者向右，一共走了10步。所以一共有 $2^{10}=1024$ 条不同的样本路径。 

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %2
设 $\{M_n,n\ge 0\}$ 是一维随机游动，设每步向右的概率为 $p=0.7$. 设 $N=100$. 求概率 $\mathbb{P}(M_{N}\ge 50)$. 
\begin{enumerate}
\item[A.]  0.1376.  
\item[B.]  0.2376. 
\item[C.]  0.3376. 
\item[D.]  0.4376. 
\end{enumerate}

{\color{red}
解答：A. 由条件可知 
\begin{eqnarray*}
\mathbb{E}(M_N) &=& n(p-q) = 100\times (0.7-0.3) = 40, \\ 
\text{Var}(M_N) &=& 4npq = 4\times 100\times 0.7 \times 0.3 = 84. 
\end{eqnarray*}
由中心极限定理，可得 
\begin{eqnarray*}
\mathbb{P} ( M_N \ge 50 ) 
= \mathbb{P} \left( \frac{M_N-40}{\sqrt{84}} \ge \frac{50-40}{\sqrt{84}} \right) \approx 1- \Phi(1.0911) = 1- 0.8624 = 0.1376. 
\end{eqnarray*} 


}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %3
考虑随机游动 $\{M_n, n\in [0,5]\}$, 事件域 $\mathcal{F}_2:=\sigma(M_0,M_1,M_2)$ 所包含的事件共有多少个？
\begin{enumerate}
\item[A.]  4.
\item[B.]  8.
\item[C.]  16.
\item[D.]  64. 
\end{enumerate}

{\color{red}
解答：C. 样本路径全体组成的集合为 
\begin{eqnarray*}
\Omega = \{ (0,1,2), (0,1,0), (0,-1,0), (0,-1,-2)\}. 
\end{eqnarray*}
所求事件域为这个集合的幂集，故有元素个数为 $2^4=16$ 个。 

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %4
设有概率空间 $(\Omega,\mathcal{F},P)$, 设事件 $A\in\mathcal{F}$, 设 $x$ 是一个实数。则与示性随机变量 $\mathds{1}_A$ 有关的事件 $\{\mathds{1}_A\ge x\}$ 不可能取到的是下述哪个事件？
\begin{enumerate}
\item[A.]  $\varnothing$.
\item[B.]  $\Omega$. 
\item[C.]  $A$. 
\item[D.]  $\bar{A}$. 
\end{enumerate}

{\color{red}
解答：D. 根据示性随机变量的定义，可知 $\{\mathds{1}_A\ge 0\}=\Omega$, $\{\mathds{1}_A\ge 1\}=A$, $\{\mathds{1}_A\ge 2\}=\varnothing$.  

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %5
关于条件期望的性质，下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  设事件域 $\mathcal{G}=\{\varnothing,\Omega\}$, 则有 $\mathbb{E}[X\,|\,\mathcal{G}]=\mathbb{E}[X]$.
\item[B.]  设 $X$ 是 $\mathcal{G}$-可测的，则有 $\mathbb{E}[X\,|\,\mathcal{G}]=X$.
\item[C.]  设有事件域 $\mathcal{G}\subseteq \mathcal{H}$, 则有 $\mathbb{E}[\mathbb{E}(X\,|\,\mathcal{H})\,|\,\mathcal{G}]=\mathbb{E}[X\,|\,\mathcal{H}]$.
\item[D.]  设 $Y$ 是 $\mathcal{G}$-可测的，则有 $\mathbb{E}[XY\,|\,\mathcal{G}]=Y\mathbb{E}[X\,|\,\mathcal{G}]$.
\end{enumerate}

{\color{red}
解答：C. 若有 $\mathcal{G}\subseteq \mathcal{H}$, 则有 $\mathbb{E}[\mathbb{E}(X\,|\,\mathcal{H})\,|\,\mathcal{G}]=\mathbb{E}[X\,|\,\mathcal{G}]$.

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %6
设 $\{W(t);t\ge 0\}$ 是标准布朗运动，则概率 $P[W(10)>3]$ 的值等于多少？
\begin{enumerate}
\item[A.]  0.1714. 
\item[B.]  0.2714. 
\item[C.]  0.3714. 
\item[D.]  0.4714. 
\end{enumerate}

{\color{red}
解答：A. 随机变量 $W(10)\sim N(0, 10)$, 是均值为零，方差为10的正态分布。所以
$$ P[W(10)>3] = P\left(\frac{W(10)}{\sqrt{10}} > \frac{3}{\sqrt{10}} \right) 
= 1 - \Phi \left( \frac{3}{\sqrt{10}} \right) = 0.1714.$$

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %7
函数 $\{f(t),0\le t\le T \}$ 的二次变差定义为
$ Q = \lim\sum\limits_{k=0}^{n-1} \Big{[} f(t_{k+1})-f(t_k) \Big{]}^2$, 
其中极限对应于划分 $0=t_0<t_1<\cdots<t_n=T$ 不断加细的过程。
求函数 $\{ f(t)=\sin(t), 0\le t\le 2\pi \}$ 的二次变差。

\begin{enumerate}
\item[A.]  $0$.
\item[B.]  $2$.
\item[C.]  $4$.
\item[D.]  $4\pi$. 
\end{enumerate}

{\color{red}
解答：A. 这是可微函数，所以二次变差为零。

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %8
关于风险的市场价格，下述说法中，不正确的是哪个？
\begin{enumerate}
\item[A.]  风险的市场价格定义为 $\lambda = \frac{\mu -r}{\sigma}$. 
\item[B.]  它是单位风险的超额回报率。
\item[C.]  风险的市场价格大于零，表明投资者冒了风险，希望平均回报率要高于无风险利率。
\item[D.]  在任何市场，每个可交易的资产都有相同的风险市场价格。
\end{enumerate}

{\color{red}
解答：D. 在完备的市场，每个可交易的资产都有相同的风险市场价格，否则会存在套利机会。

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item %9
下述求解几何布朗运动模型 $\frac{dS(t)}{S(t)}=rdt+\sigma dW(t)$ 的步骤中，不正确的是哪个？
\begin{enumerate}
\item[A.]  根据 Ito 公式，可得 $d \ln S(t) = \frac{1}{S(t)}dS(t) - \frac{1}{2S^2(t)}[dS(t)]^2$. 
\item[B.]  根据几何布朗运动模型，可得 $d \ln S(t) = rdt+\sigma dW(t) - \frac{1}{2}[rdt+\sigma dW(t)]^2$. 
\item[C.]  根据 $dW(t)^2=dt$ 并忽略高阶项，可得 $d \ln S(t) = rdt+\sigma dW(t) - \frac{1}{2}\sigma^2 dt$. 
\item[D.]  两边积分，可得 $ \ln S(t) = \ln S_0 + (r-\frac{1}{2}\sigma^2)t + \sigma W(t)$. 
\item[E.]  消去对数运算，可得 $ S(t) = S_0 +\exp\left[ (r-\frac{1}{2}\sigma^2)t + \sigma W(t) \right] $. 
\end{enumerate}

{\color{red}
解答：E. 消去对数运算，结果是初始值乘以指数函数。

}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}



